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The image shows a full-wave rectifier circuit with two diodes (D1, D2), a transformer with a turn ratio of 18:1, a capacitor \( C \), and a load resistor \( R_L \). The power supply voltage is \( V_s = 220 \, V \). The problem is split into two parts:

### Problem (a):
- Given: 
  - \( R_L = 10 \, \Omega \),
  - \( C_1 = 1000 \, \mu F \),
  - \( C_2 = 2000 \, \mu F \).

The task is to determine the output voltage \( V_{LDC} \).

### Problem (b):
- Given:
  - \( C = 1000 \, \mu F \),
  - \( R_L = 10 \, \Omega \),
  - \( R_L = 100 \, \Omega \).

The task is to calculate the ripple factor \( r \) in percentage.

### Step-by-Step Solution Outline:

#### For part (a):

1. **Peak Secondary Voltage Calculation**:
   The transformer reduces the input voltage to the secondary side by a factor of 18:
   \[
   V_{peak} = \frac{V_s}{18 \sqrt{2}} = \frac{220}{18 \sqrt{2}} \, V
   \]
   This gives the peak value of the rectified voltage before smoothing.

2. **DC Output Voltage**:
   The capacitor smooths the rectified signal. The approximate DC output voltage \( V_{LDC} \) is:
   \[
   V_{LDC} \approx V_{peak} - V_{diode-drop}
   \]
   where \( V_{diode-drop} \approx 0.7 \, V \) for each diode.

#### For part (b):

1. **Ripple Voltage**:
   The ripple voltage \( V_{ripple} \) for a full-wave rectifier is given by:
   \[
   V_{ripple} = \frac{I_{load}}{f C}
   \]
   where:
   - \( I_{load} = \frac{V_{LDC}}{R_L} \),
   - \( f \) is the frequency of the ripple (which is twice the mains frequency for full-wave rectification, typically \( 2 \times 50 = 100 \, Hz \)).

2. **Ripple Factor**:
   The ripple factor \( r \) is calculated as:
   \[
   r = \frac{V_{ripple}}{V_{LDC}} \times 100 \%
   \]

Would you like the full numeric calculations for each part?

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### Related Questions:
1. What effect does increasing the capacitance have on the ripple factor?
2. How does the load resistance impact the output voltage?
3. Why is the frequency of ripple in a full-wave rectifier twice the mains frequency?
4. What would happen to \( V_{LDC} \) if the diode drop were larger (e.g., with different diodes)?
5. How do you choose an appropriate capacitor for a rectifier circuit?

### Tip:
Larger capacitance in the smoothing capacitor reduces the ripple voltage but increases charging current, which may affect diode stress.

Determine the output voltage and ripple factor for a full-wave rectifier circuit given specific component values.

Explore the step-by-step solution to determine the output voltage and ripple factor in a full-wave rectifier circuit. Understand key concepts in AC to DC conversion, capacitor smoothing, and rectifier circuit analysis. Suitable for undergraduate electronics engineering students.

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